Trig: Solving 6cosA+3=2sinA with Given Conditions

[DERRAN]

Homework Statement

Given: sec$$\theta$$=$$\sqrt{10}$$ where 0< $$\theta$$ <90

and $$\sqrt{10}$$sin(A-$$\theta$$)=sinA-3cosA



Determine the solution of
6cosA +3 = 2sinA


for A $$\in$$ [-180; 180], rounded off to one decimal digit.

Homework Equations

The Attempt at a Solution

3=2sinA - 6cosA

3=2(sinA-3cosA)

$$\frac{3}{2}$$=sinA-3cosA

$$\frac{3}{2}$$=$$\sqrt{10}$$sin(A-$$\theta$$)

Now I can't get rid of the theta

 

[rock.freak667]

Homework Statement

Given: sec$$\theta$$=$$\sqrt{10}$$ where 0< $$\theta$$ <90

and $$\sqrt{10}$$sin(A-$$\theta$$)=sinA-3cosA



Determine the solution of
6cosA +3 = 2sinA


for A $$\in$$ [-180; 180], rounded off to one decimal digit.

Homework Equations

The Attempt at a Solution

3=2sinA - 6cosA

3=2(sinA-3cosA)

$$\frac{3}{2}$$=sinA-3cosA

$$\frac{3}{2}$$=$$\sqrt{10}$$sin(A-$$\theta$$)

Now I can't get rid of the theta

if $sec \theta = \sqrt{10}$ that means $cos \theta = ?$

 

[DERRAN]

cos$$\theta$$=1/$$\sqrt{10}$$
but that still doen't help with getting rid of the theta over here.

3/2=$$\sqrt{10}$$sin(A-$$\theta$$)

 

[rock.freak667]

cos$$\theta$$=1/$$\sqrt{10}$$
but that still doen't help with getting rid of the theta over here.

3/2=$$\sqrt{10}$$sin(A-$$\theta$$)

erm...


$$cos \theta = \frac{1}{\sqrt{10}}$$

$\theta$ is what then in the range $0< \theta < 90$ ?

 

[DERRAN]

erm...


$$cos \theta = \frac{1}{\sqrt{10}}$$

$\theta$ is what then in the range $0< \theta < 90$ ?

got it Thank you.